mass transfer chapter 3 diffusion in concentrated … flow ! note: this convection also transports...

11. October 2017

Chapter 3Diffusion in Concentrated Solutions

Mass Transfer

Mass Transfer – Diffusion in Concentrated Solutions 3-2

3. DIFFUSION IN CONCENTRATED SOLUTIONS

Diffusion causes convection in fluidsConvective flow occurs because of pressure gradients (most common) or temperature differences (buoyancy or free or natural convection). However even in isothermal and isobaric systems, convection can occur due to diffusion. This is called “diffusion-induced convection”

In general, diffusion and convection always occur together in fluids.

Maxwell (1860): “Mass transfer is due partly to the motion of translation and partly that of agitation.”

3.1 Theory

Mass Transfer – Diffusion in Concentrated Solutions

A first example to illustrate diffusion-induced convection: Evaporation of Benzene

3-3

At 6°C the benzene vapor is dilute and evaporation is limited by diffusion.

At 80.1°C benzene boils (p = 1atm).Evaporation is controlled by convection.

At 60°C an intermediate case occurs in which both diffusion and convection are important.


Analysis of the case at 60°C:

3-4

Concentration of benzene (species 1) and air (species 2) at z = h:

c1,h = c1,∞ (→ 0); c2,h → max

p1,h = p1,∞ (→ 0); p2,h → p

(air blows away and dilutes benzene vapor).

At z = 0:

c1,0 = f (T,p) > c1,h and c2,0 < c2,h

p1,0 = f (T,p) > p1,h and p2,0 < p2,h


The difference in concentrations (or partial pressures) between 0 and h gives rise to upward diffusion of benzene from the liquid surface and downward diffusion of air.

Since the benzene surface is considered impermeable to air, a convective upward-pointing flux must compensate the downward diffusive flux of air!

This is the Stefan flow !

NOTE: This convection also transports benzene molecules!


Determine concentration profile and flux for the 60°C case

3-6

Mass balance for species i at steady state:

)zz(J)zz(J)z(J)z(J diff,iconv,idiff,iconv,i ∆++∆+=+

Diffusive flux:

dzdcADJ diffi ⋅−=,

Convective flux with velocity u:

AucJ iconv,i ⋅⋅=


Divide by A Δz and Δz → 0: 02

2=

∂∂

⋅+

∂∂

⋅−zcD

zcu ii

For practicability we transform molar concentrations into mole fractions:

( ) cc

VccVc

nnn

y 1

21

1

2,M1,M

1,M1 =

⋅+⋅

=+

=

(1)

If c is given as a mass concentration the transformation into mole fractions is:

1

11 M

Mcy⋅ρ⋅

= with the “average molar mass” i

ki MyM ∑ ⋅= 1

where nM,i is the number of moles of species i.


So (1) can be rewritten as: 02

2=

∂∂

⋅+

∂∂

⋅−⋅zyD

zyuc ii (2)

Integration of eq. (2) for species 1 (benzene):

CzyDyuc =

∂∂

⋅−⋅⋅ 11 (3)

The two terms are the convective and diffusive molar flux densities. The sum of the molar fluxes is constant.

Integration of eq. (2) for species 2 (air) with 12 1 yy −=

( ) CzyDy1uc 1

1 ′=

∂∂

⋅+−⋅⋅ (4)


Since air does not accumulate and cannot penetrate the benzene surface, the total (overall) molar flux of air is zero. Thus 0=′C

Eq. (4) is rewritten as: ( )

∂∂

−=−zy

uDy 1

11 (5)

Integration of (5) with B.C. y1 = y1,0 at z = 0:

( ) ( ) zDu

eyy ⋅−=− 0,11 11 (6)

Determine velocity u with B.C. y1 = y1,h at z = h:

−−

⋅=0,1

,1

11

lnyy

hDu h

(7)


Inserting eq. (7) into (6) gives the concentration profile:

( )( )

hz

h

yy

yy

−−

=−−

0,1

,1

0,1

1

11

11 (8)

With the concentration profile and eq. (3) we can obtain the total molar flux:

−−

⋅⋅

=0,1

,1

11

lnyy

hDcj h

tot (9)

This type of approach leads to the Stefan-Maxwell equations for multi-component diffusion in concentrated solutions.


Separating Convection from Diffusion – Cussler’s Approach

3-11

Now, Cussler approaches this topic slightly different but gets to the same results:

Again, assume that the two transport effects are additive:

+

=

convection by

dtransporte massdiffusion by

dtransporte massdtransporte mass total

If the total mass flux is n1, the mass transported per area per time relative to fixed coordinates:

111 vcn =

where v1 is the average solute velocity (velocity due to convection and superimposed diffusion).


The total average solute velocity can be split into one part due to diffusion and one due to convection, called reference velocity va:

( )

convection

a1

fluxdiffusive

a1

a1

a111 vcjvcvvcn +=+−=

The art is to select va in such a way that the convection term is simplified or ideally: va=0.

For example va is the velocity of the solvent because the solvent is usually in excess so its transfer is minimal (in other words the difference in solvent concentration is too small across the solution). That way we eliminate convection and deal with a SIMPLER problem.


Two-bulb apparatus (Diaphragm-cell) for understanding different definitions of reference velocities.

Volume average velocity = 0Molar average velocity = 0Mass average velocity ≠ 0

Volume average velocity = 0Molar average velocity ≠ 0Mass average velocity ≈ 0In

itial

cent

er o

f mol

es

Fina

l

cent

er o

f mol

es


For gases (e.g. H2 and N2) at equal T and p the number of moles is always the same in both sides. Similarly the volume in both sides is the same. As a result, the

v0 = 0 volume average velocity

v* = 0 molar average velocity

v ≠ 0 mass average velocity, because the masses of N2 and H2 are different. As a result, as time goes by the center of mass in the two-bulb apparatus moves away from the bulb containing N2 initially. Thus the mass average velocity v is not zero.


For liquids: The volume is nearly always constant.v0 = 0 volume average velocityv = 0 mass average velocity. This is usually correct as liquid

densities differ little.e.g. ρH2O=1 g/cm3

ρGlycerol=1.1 g/cm3

However, the molar concentration is usually quite different following large differences in molecular weight.

e.g. MWH2O=18 g/mol andMWGlycerol=92 g/mol

So v*≠0 molar average velocity for liquids.

In conclusion: For gases use as reference velocity va the v0 or v*, while for liquids use v0 or v.


Table legend:

ω i : mass fraction of species iy i : mole fraction of species i

ii Vc : volume fraction of species i, iVwhere is the partial specific volume of species.

Precisely: Partial specific volume:jmTpi

i mVV

,,

∂∂

=

Partial molar volume:j,Mn,T,pi,M

i nVV

∂∂

=

The partial specific or molar volume expresses how much a volume changes upon addition of a certain mass or number of moles of a given component.


For ideal gases (pV = nRT),

( )c1

pTR

np

RTnn

nVV

2,M

2,M

n,T,p

1,M

2,M1,M

n,T,p1,M1 =

⋅=

∂

+∂

=

∂∂

=

iV can be expressed as:


3.2 Examples for Parallel Diffusion and Convection

3-19

Example 3.2.1: Fast diffusion through a stagnant film

Goal: Calculate the flux and the concentration profile

Now both diffusion and convection are important!

Remember that at intermediate temperatures both diffusion and convection affect the evaporation of benzene (or any other solute).


1. Step: Mass balance

∆+

−

=

∆ zzatout

dtransportesolutezatin

dtransportesolutezAvolumein

daccumulatesolute

( ) zz1z11 |An|AnczAt ∆+−=∆∂∂

Divide by A∆z and as volume 0

znc

t1

1 ∂∂

−=∂∂ At steady state:

zn0 1∂∂

=


Now the flux is affected by both diffusion and convection. For simplicity we choose va=v0 (volume average velocity)

)vVcvVc(cdzdcDvcjn 2221111

10111 ++−=+=

2. Step: Choose and simplify mass transport equation

Note that n1=c1v1 and n2=c2v2

The total average flux of the solvent (air) is zero (it seems to be stagnant), since it cannot penetrate into the liquid phase and does not accumulate.

Therefore n2=0 and v2=0.


So 1111

1 nVcdzdcDn +−=

If the vapor is an ideal gas, then ( ) 111

21

11 Vc

cc

VccVcy ==⋅+

⋅=

dzdcD)Vc1(n 1111 −=−

dzdyDc)y1(n 1

11 −=− (1)

3. Step: Boundary conditions z=0: y1=y10

z=L: y1=y1L

(2)

(3)

or

and


Solve eqn. (1) subject to BC’s to determine n1

−−

=10

11 y1

y1lnDcn

(4)

Note that doubling the concentration difference DOES NOT double the flux, as in dilute systems.

−−

⋅⋅

==0,1

,1

11

lnyy

hDcCj h

tot

Compare to our initial result for combined diffusion and convection of benzene (page 3-10):

The direct approach and the one using the reference velocity give the same results!

TotalFlux of benzene


−−

−−−

−= =10

1

z

10

110

y1y1ln

y1y1y1Dc

dz1dy

Dc1j

(6)Diffusive

Flux of benzene

Integrating eq. (1) also for z=0 to z and y1=y10 to y1 and assuming that n1 does not change with height z (which is a fair assumption here as the cross-sectional area does not change) gives:

z

10

1

10

1y1y1

y1y1

−−

=−−

(5)Concentration profile

With (5) and Fick's law we can determine the diffusive flux:


Now does this result (eqn. 5,6) reduce to that for dilute solutions?

Expansion into series for small y (dilute system small conc. y):

( ) ( ) ( )( ) +

++++⋅=± − 32a y

!32a1aay

!21aaya1y1

(7)

(8)

(9)

( ) ( ) ( )( ) ya1y!3

2a1aay!2

1aaya1y1 32a ⋅±≈+−−

±−

+⋅±=±

Here for z=l, thus a=1: y1y1

1+≈

−

( ) y3y

2yyy1ln

32

−≈

+++−=−

0 0


Let´s apply eqn. (8) to eqn. (5)

( )( )[ ] [ ]

z101

z101

10

1 yy1y1y1y1y1

+−=+−=−−

( )

z

101101 yy1)y1(y1 +−−=−

( ) ( ) 1101010110 yyzy1yzyz1y1 −+

−≈+−⋅−≈

If we rearrange and multiply both sides of eqn. (10) with c

(10)

)cc(zcc 101101 −+=

(11)


Likewise for the flux from eqn. (4)

[ ]

[ ] )cc(DyyDc

)y1ln()y1ln(Dcn

1101019Eq

1011

−=+−=

−−−=

Eqn. (11) and (12) are identical to the dilute limit ones!

(12)


Example 3.2.2: Calculate the error associated with the neglect of diffusion-driven convection when estimating the evaporation rate of benzene @ 6°C and @ 60°C.a) At 6°C the saturation vapor pressure is p1(sat) = 37 mmHg

Mole fraction 049.076037

p)sat(p

ccy 11

10 ====

Total flux at steady-state for concentrated solution:

cD05.0

049.0101lncD

y1y1lncDn10

11 =

−−

=

−−

=

Total flux for dilute solution:

( ) ( )

cD049.00049.0cDyycDjn 11011 =−=−== Only 2% error!


b) At 60°C the saturation vapor pressure is p1(sat) = 395 mmHg

760395y10 =Mole fraction

Concentrated solution:

cD73.0

760/395101lncD

y1y1lncDn10

11 =

−

−=

−−

=

Dilute solution:

( )

cD52.00

760395cDyycDjn 11011 =

−=−==

There is 40% error !!


−−

=10

11 y1

y1lnDcn

z

10

1

10

1y1y1

y1y1

−−

=−−

(6):

Physical picture for species 1:

(4):

(5):y1

−−

−−−

−=

=10

1

z

10

110

y1y1ln

y1y1y1Dc

dz1dy

Dc1j


Example 3.2.3: Hydrogen production by catalytic cracking of CH4

Methane gas is cracked at the surface of a solid catalyst forming hydrogen and a solid carbon deposit.

Goal: Total methane (molar) flux per unit area at steady state?

n1 Catalyst surface

n2

z 0

CH4

2H2 Carbon deposit

CH4(g) → C(s) + 2 H2(g)


Choose and simplify mass transport equation:

Note: For processes with chemical reactions, it is best to use the molar flux and the molar average velocity!

Thus, from Table 3.2.1: *1

1*111 vc

dzdcDVcjn ⋅+−=⋅+= (1)

withc

nnvyvyv 212211

* +=+= (2)

Now 1 mole of CH4 gives 2 moles of H2, flowing in the opposite direction. Therefore,

12 2 nn ⋅−= in eq. (2):cnv 1* −= So: c

ncdzdcDn 1

11

1 ⋅−−=


Use molar fractions ccy 1

1 =

111

1 nydzdycDn ⋅−⋅⋅−= ( )

dzdycDyn 1

11 1 ⋅⋅−=+⋅→ (3)

B.C.: z = 0: y1 = y1,0 = 0 (due to decomposition)z = L: y1 = y1,L (some measured conc. at L)

Integration of (3) subject to B.C.s yields: ( )LyL

Dcn 11 1ln +⋅−=

Or the general form if y1,0 ≠ 0:

++

⋅−=0,1

,11 1

1ln

yy

LDcn L


Example 3.2.4: Fast Diffusion into Semi-Infinite SlabA volatile liquid solute evaporates into a long capillary.

Initially the capillary contains no solute. As the solute evaporates the interface between the vapor and the liquid solute drops.

Goal: Calculate the solute evaporation rate accounting for diffusion-induced convection and the effect of moving interface.

There is no solvent (air) flow across the capillary, blowing the solute away. As a result, the solute accumulates in the capillary.


Mass balance:

−

=

∆ outtransport

soluteintransport

solutezAin

ionaccummulatsolute

( ) ( ) ( ) zz1z11 nAnAzAct ∆+−=∆

∂∂

Divide by A ∆z and as ∆z 0: zn

tc 11

∂

∂−=

∂

∂(1)

Choose and simplify mass transport equation:

n1 = j1+c1 v0 22112221110 VnVnvVcvVcv +=+=with

012

12

1 vczz

cD

tc

∂∂

−∂

∂=

∂

∂In (1): (2)

Now find an expression for v0 !


In the unsteady case, the solvent flux varies with position and time but the solvent gas does not dissolve in the liquid, thus at the interface (z=0): n2 = 0.

( ) 0z1

0z111 zc

DnVc1 == ∂

∂−=− (3)

( )0z111

0z

122111

11 Vnc

zcDVnVnc

zcDn

==

+∂∂

−=++∂∂

−=

and ( )0z

11

1

0z1 Vc1zcD

n

=

=

−∂∂

−=

( ) constVVc1zcD

Vnv 1

0z

11

1

110 =

−∂∂

−==

=

So,


t = 0 ∀ z > 0 c1 = 0t > 0 z = 0 c1 = c1(sat)

z = ∞ c1 = 0

Boundary conditions:

Define combined variable (as in the dilute case): tDz

⋅⋅=ζ

4

with B.C. ζ = 0 c1 = c1(sat)ζ = ∞ c1 = 0

zc

Vc1z

cVD

z

cD

tc 1

0z

11

11

21

21

∂

∂

−

∂

∂

+∂

∂=

∂

∂

=

(4)Insert in (2):


where

0

11

11

Vc1

cV

21

=ζ

−

ζ∂

∂

−=Φ (6)

In eq. (5) Φ is a dimensionless velocity characterizing the convection by diffusion and the movement of the interface. Note that if Φ = 0 the problem reduces to that of diffusion in dilute concentrations !!

Eqn. (5) is integrated to give: ( ) [ ]21 )(expttanconsc

Φ−ζ−=ζ∂

∂

( ) 0d

dc2

d

cd 121

2=

ζΦ−ζ+

ζ(5)(4) becomes:


2nd integration and insertion of B.C.:

( )( )Φ+Φ−ζ−

=erf

erfsatc

c1

1)(1

1(7)

eqn. (6) (7):( )( ) [ ]

1

211experf1

11)sat(cV−

ΦΦΦ+π+=

Calculate now also the interfacial flux (see eq. 3)

0z

11

14.eqn

0z11 Vc1z

c

DnN

=

=

−

∂

∂

−= =[ ]( )Φ+Φ−

−π=

erf1exp

)sat(cV11)sat(ct/D

2

11itlimdilute

1

mass transfer chapter 3 diffusion in concentrated … flow ! note: this convection also transports...

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